Three Independent Tests for a Measure
A set function can resemble a measure in some respects and still fail to be one. A useful diagnosis starts by separating the requirements rather than relying on a general impression that the function “adds up correctly.” In particular, nonnegativity, the empty-set condition, and countable additivity are distinct requirements. Their logical relationships also depend on whether the set function is allowed to take the value \(+\infty\).
The underlying collection must be a sigma-algebra, and the set function must be defined on that whole collection. The examples below use power sets, which are sigma-algebras. We will identify a failure by giving the precise set or disjoint sequence that contradicts an axiom. A test on one sequence can disprove countable additivity; passing a few tests, by contrast, does not establish it.
Consequences of Countable Additivity
Proof. Let \(E_1,\ldots,E_k\in\mathcal{M}\) be pairwise disjoint. Extend this to a sequence by setting \(E_n=\varnothing\) for \(n>k\). The union of the sequence is \(\bigcup_{j=1}^k E_j\), and countable additivity, together with \(\mu(\varnothing)=0\), gives \(\mu(\bigcup_{j=1}^k E_j)=\sum_{j=1}^k\mu(E_j)\). For monotonicity, \(F\) is the disjoint union of \(E\) and \(F\setminus E\). Finite additivity gives \(\mu(F)=\mu(E)+\mu(F\setminus E)\), and the second term is nonnegative. Hence \(\mu(E)\le\mu(F)\), including when \(\mu(F)=+\infty\). \(\square\)
These consequences are useful diagnostic tools. If a candidate fails finite additivity for two disjoint sets, it cannot be countably additive. But finite additivity is only a consequence of the measure axioms; verifying it on finite families does not by itself establish countable additivity.
Worked Example: A Signed Set Function That Is Additive
Let \(X=\{a,b\}\), \(\mathcal{M}=\mathcal{P}(X)\), and define the real-valued set function \(\nu(A)=-\chi_A(a)+2\chi_A(b)\). Check which measure axiom fails.
Its values on all four sets are \(\nu(\varnothing)=0\), \(\nu(\{a\})=-1\), \(\nu(\{b\})=2\), and \(\nu(X)=1\). Thus \(\nu(\{a\})=-1<0\), so the nonnegativity requirement fails.
In fact, the empty-set condition and countable additivity both hold. For any pairwise disjoint sequence \((E_n)\) of subsets of \(X\), each of \(a\) and \(b\) belongs to at most one \(E_n\). Therefore the series \(\sum_n\nu(E_n)\) has at most two nonzero terms, and \(\sum_n\nu(E_n)=-\chi_{\cup_n E_n}(a)+2\chi_{\cup_n E_n}(b)=\nu(\cup_n E_n)\). For instance, the disjoint sets \(\{a\}\) and \(\{b\}\) give \(\nu(X)=1=(-1)+2\). The negative singleton value, not an additivity error, is the reason \(\nu\) is not a measure.
The Empty-Set Condition Is Separate
It is tempting to think that countable additivity automatically forces the value at the empty set to be zero. With values in \([0,\infty]\), that claim is false: an infinite value can satisfy the additivity equation even for a sequence consisting entirely of empty sets.
Worked Example: Countable Additivity Without the Empty-Set Axiom
Take \(X=\{a\}\), \(\mathcal{M}=\mathcal{P}(X)\), and define \(\nu(A)=+\infty\) for every \(A\in\mathcal{M}\). This set function is nonnegative, and \(\nu(\varnothing)=+\infty\ne0\).
Nevertheless, it is countably additive. For every sequence \((E_n)\) of measurable sets, the union \(\bigcup_nE_n\) belongs to \(\mathcal{M}\), so \(\nu(\bigcup_nE_n)=+\infty\). Also \(\nu(E_n)=+\infty\) for each \(n\), and consequently \(\sum_n\nu(E_n)=+\infty\). The two sides agree, in particular for every pairwise disjoint sequence. The failure is exactly the empty-set axiom: countable additivity and nonnegativity do not rule out this set function.
By contrast, if a countably additive set function takes only finite nonnegative values, then the empty-set condition does follow. Indeed, let \(c=\nu(\varnothing)\). Applying countable additivity to the sequence \(E_n=\varnothing\) gives \(c=\sum_{n=1}^{\infty}c\). If \(c>0\), the sum on the right is \(+\infty\), contradicting the finiteness of \(c\). Thus \(c=0\). This argument does not apply when \(c=+\infty\), as the example demonstrates.
This distinction matters when checking definitions: the empty-set axiom is stated separately because measures are allowed to take infinite values. It also prevents an invalid shortcut in which countable additivity is checked and the empty-set condition is simply assumed to follow.
Testing Countable Additivity Directly
Worked Example: A Nonnegative Set Function That Is Not Additive
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and define \(\nu(A)=0\) when \(A\) is finite, and \(\nu(A)=1\) when \(A\) is infinite. This set function takes values in \([0,\infty)\) and \(\nu(\varnothing)=0\).
Use the pairwise disjoint sequence \(E_n=\{n\}\). Every \(E_n\) is finite, so \(\nu(E_n)=0\), and \(\sum_{n=1}^{\infty}\nu(E_n)=\sum_{n=1}^{\infty}0=0\). But \(\bigcup_n E_n=\mathbb{N}\) is infinite, so \(\nu(\bigcup_nE_n)=1\). Thus the required equation would say \(1=0\), and countable additivity fails.
This candidate also fails finite additivity, so the example does not show a failure unique to infinite sequences. For the disjoint sets \(E_1=\{1\}\) and \(E_2=\{2\}\), their union is finite and has value \(0\), whereas \(\nu(E_1)+\nu(E_2)=0+0=0\); that particular test passes. Instead, take the disjoint infinite sets \(A=\{2,4,6,\ldots\}\) and \(B=\{1,3,5,\ldots\}\). Then \(\nu(A\cup B)=\nu(\mathbb{N})=1\), while \(\nu(A)+\nu(B)=1+1=2\). This finite test already rules out a measure. The singleton sequence supplies a separate, explicit countable-additivity contradiction.
When a candidate is defined on an infinite sigma-algebra, a good diagnostic is to look for a natural disjoint decomposition of a set whose value is easy to calculate. Then calculate both sides of the countable-additivity equation, including the full series on the right. Do not infer the sum from a finite partial sum: a sequence of positive terms can have an infinite sum, and the measure axiom allows the value \(+\infty\).
Confirm that the function is defined on every member of a sigma-algebra and takes values in \([0,\infty]\). A negative value already rules out a measure.
Evaluate the function at \(\varnothing\). The required value is exactly \(0\), not merely a finite number or a value that behaves well on other sets.
Choose pairwise disjoint measurable sets, compute the value of their union, and compare it with the entire nonnegative series of their individual values.
A Common Diagnostic Mistake
A common mistake is to claim that any failure of the measure definition must be a failure of additivity. The signed example shows otherwise: its countable-additivity equation holds, but its value on \(\{a\}\) is negative. The constant-\(+\infty\) example shows the converse kind of logical separation: countable additivity and nonnegativity can hold while the empty-set condition fails.
Another mistake is to treat finite additivity as a substitute for countable additivity. The proposition proves that a measure must be finitely additive, so a finite-additivity counterexample is decisive. But passing finite tests does not establish the countable equation. For any proposed countable test, the full union and the full series must be compared. The singleton decomposition of \(\mathbb{N}\) in the last example gives exactly such a comparison: the union has value \(1\), while the series is \(0+0+\cdots=0\).
Keeping the axioms separate makes a diagnosis precise. When a candidate fails, state the particular axiom and exhibit the value or disjoint sequence that contradicts it. When a candidate passes some axioms, say so as well; this prevents an unrelated failure from being mistakenly attributed to additivity.
Check Your Understanding
For each question, identify the relevant set function or axiom and give the calculation that supports your answer.
- On \(X=\{a,b\}\), define \(\nu(A)=3\chi_A(a)-\chi_A(b)\). Which measure axiom fails? Does the empty-set condition hold?
- Why does the constant set function \(\nu(A)=+\infty\) satisfy countable additivity on a nonempty power set, even for a sequence of empty sets?
- For \(\nu(A)=0\) when \(A\subseteq\mathbb{N}\) is finite and \(\nu(A)=1\) when \(A\) is infinite, calculate both sides of countable additivity for \(E_n=\{n\}\).
- Prove that every measure is finitely additive on pairwise disjoint finite families by extending the family with empty sets.
- If a countably additive set function takes only finite nonnegative values, show that its value at the empty set must be zero.